Download Full text

ADVANCED PROBLEMS AND SOLUTIONS
Edited by
Florian Luca
Please send all communications concerning ADVANCED PROBLEMS AND SOLUTIONS to FLORIAN LUCA, IMATE, UNAM, AP. POSTAL 61-3 (XANGARI), CP 58 089,
MORELIA, MICHOACAN, MEXICO, or by e-mail at [email protected] as files of the
type tex, dvi, ps, doc, html, pdf, etc. This department especially welcomes problems believed to
be new or extending old results. Proposers should submit solutions or other information that
will assist the editor. To facilitate their consideration, all solutions sent by regular mail should
be submitted on separate signed sheets within two months after publication of the problems.
PROBLEMS PROPOSED IN THIS ISSUE
H-639 Proposed by H.-J. Seiffert, Berlin, Germany
The sequences of the Fibonacci and Lucas polynomials are defined by
F0 (x) = 0, F1 (x) = 1, and Fn+1 (x) = xFn (x) + Fn−1 (x) for n ≥ 1,
L0 (x) = 2, L1 (x) = x, and Ln+1 (x) = xLn (x) + Ln−1 (x) for n ≥ 1,
respectively. Prove that, for all non-zero complex numbers x and all positive integers n,
(a)
2n−1
X k=0
4n − 1 − k 4n−1−2k k
2
x Fk (x) = x2n−1 L2n−1 (x)F2n (4/x),
k
(b)
2n−1
X k=0
4n − 1 − k 4n−1−2k k
2
x Lk (x) = x2n−1 (x2 + 4)F2n−1 (x)F2n (4/x),
k
(c)
2n X
4n + 1 − k
k=0
k
24n+2−2k xk Fk (x) = x2n+1 F2n (x)L2n+1 (4/x),
(d)
2n X
4n + 1 − k
k=0
k
24n+2−2k xk Lk (x) = x2n+1 L2n (x)L2n+1 (4/x).
187
ADVANCED PROBLEMS AND SOLUTIONS
H-640 Proposed by Jayantibhai M. Patel, Ahmedabad, India
For any positive integer n ≥ 2, prove that the value of the following determinant
−L2n
F2n
2
Ln
2F2n
2
Ln
F2n
L2n
−3(3Fn2 + 2(−1)n ) F2n
F2n
−L2n
2
2Fn
2F2n
F2n
L2n
2F2n
2Fn2
2F2n
−6Fn+1 Fn−1
2F2n
L2n F2n L2n 2F2n −L2n
is (L2n + L2n )5 .
H-641 Proposed by Emeric Deutsch, Polytechnic University, Brooklyn, NY
A composition of n is an ordered sequence of positive integers having sum equal to n. The
terms of the sequence are called the parts of n. It is known that the number of compositions
of n is 2n−1 and the number of compositions of n with exactly k parts is equal to n−1
k−1 . Here,
we consider a slightly modified concept assuming that there are two kinds of 1. Find:
(i) The number an of compositions of n (for example, a2 = 5 because we have
(2), (1, 1), (1, 10 ), (10 , 1), and (10 , 10 ));
(ii) the number cn,k of compositions of n with exactly k parts (for example, c4,2 = 5
because we have (1, 3), (10 , 3), (3, 1), (3, 10 ), and (2, 2)).
H-642 Walther Janous, Innsbruck, Austria
Determine the limit
lim
L2
n+2
n→∞
Fn+2
−
n
X
L2 k
k=1
Fk
.
SOLUTIONS
Bounding ratios of Fibonacci numbers
H-625 Proposed by Russel Jay Hendel, Townson University, MD
(Vol. 3, no. 2, May 2005)
For an integer m > 0 let Km be the smallest positive integer such that Fn+m < Km Fn
holds for all large n. For example, K1 = 2 because Fn < Fn+1 < 2Fn holds for all large n.
Provide an explicit formula for Km .
Solution by H.-J. Seiffert
It is known (see equations (3.22) and (3.24) in [1]) that, for all integers m and n,
Fn+m + (−1)m Fn−m = Lm Fn .
188
(1)
ADVANCED PROBLEMS AND SOLUTIONS
We shall prove that, for m > 0,
(
Km =
Lm + 1,
Lm ,
if m is odd,
if m is even.
Suppose that m is odd. If n > m, then, by (1),
Lm Fn = Fn+m − Fn−m < Fn+m < Fn+m + Fn − Fn−m = (Lm + 1)Fn .
Let m be even. If n > m, then, by (1) again,
(Lm − 1)Fn = Fn+m − (Fn − Fn−m ) < Fn+m < Fn+m + Fn−m = Lm Fn .
This completes the proof of the above statement.
[1] A. F. Horadam & Bro. J. M. Mahon. “Pell and Pell-Lucas Polynomials,” The Fibonacci
Quarterly 23.1 (1985) : 7–20.
Also solved by Paul S. Bruckman and the proposer.
Pell numbers and Fibonacci polynomials
H-626 Proposed by H.-J. Seiffert, Berlin, Germany
(Vol. 43, no. 2, May 2005)
The Fibonacci polynomials are defined by F0 (x) = 0, F1 (x) = 1, and Fn+2 (x) =
xFn+1 (x) + Fn (x) for n ≥ 0. Let n be a positive integer.
a. Prove that, for all complex numbers x,
Fn+1 (x) + iFn (x) = 4
−n
n X
2n + 1
k=0
2k + 1
(x − 2i)k (x + 2i)n−k ,
where i =
√
−1.
b. Deduce the identities
−b n
2c
Pn = 2
n−1
X
k=0
46|n−2k+1
b
(−1)
(n−2k)
c
4
n
X
(n+1)
(n−2k−1)
2n − 1
−b 2 c
b
c 2n + 1
4
and Pn = 2
(−1)
,
2k + 1
2k + 1
k=0
46|n−2k
where Pn = Fn (2) is the nth Pell number.
Solution by the proposer
189
ADVANCED PROBLEMS AND SOLUTIONS
It is well-known that
F2n+1 (x) = √
x + √x2 + 4 2n+1
1
2
x2 + 4
−
x − √x2 + 4 2n+1 2
.
Applying the Binomial Theorem gives
F2n+1 (x) = 4
−n
n X
2n + 1
2k + 1
k=0
(x2 + 4)k x2n−2k .
√
√
Replacing x by ix − 2 (here, ix − 2 can be any of the at most two possible square roots of
ix − 2) and multiplying by (−i)n yields
n X
√
2n + 1
−n
(−i) F2n+1 ( ix − 2) = 4
(x − 2i)k (x + 2i)n−k .
2k + 1
n
k=0
Using the known relation F2j (y)/y = i1−j Fj (i(y 2 + 2)) with y =
Fj (−x) = (−1)j−1 Fj (x), we find
√
ix − 2 and noting that
√
√
√
(−i)n (−i)n F2n+1 ( ix − 2) = √
F2n+2 ( ix − 2) − F2n ( ix − 2) = Fn+1 (x) + iFn (x).
ix − 2
This proves the identity of part a.
√
√
Since 1 − i = 2e−iπ/4 and 1 + i = 2eiπ/4 , the identity of a with x = 2 implies that
−n/2
Pn+1 + iPn = 2
n X
2n + 1
(n − 2k)πi .
exp
4
2k + 1
k=0
Using Euler’s relation eiy = cos y + i sin y, after equating the real and imaginary parts, we find
−n/2
Pn+1 = 2
n X
2n + 1
2k + 1
k=0
An−2k ,
(1)
and
−n/2
Pn = 2
n X
2n + 1
k=0
2k + 1
190
Bn−2k ,
(2)
ADVANCED PROBLEMS AND SOLUTIONS
where Aj = cos(jπ/4) and Bj = sin(jπ/4), for j ∈ Z. Simple calculations show that
(
(−1)b(j+1)/4c 2bj/2c−j/2 ,
0,
if j 6≡ 2 (mod 4),
if j ≡ 2 (mod 4),
(
(−1)b(j−1)/4c 2bj/2c−j/2 ,
0,
if j 6≡ 0 (mod 4),
if j ≡ 0 (mod 4).
Aj =
Bj =
Now the identities of part b follows from (1) with n replaced by n − 1 and (2).
Also solved by Paul S. Bruckman.
Revisiting the Cauchy-Schwartz inequality
H-627 Proposed by Slavko Simic, Belgrade, Yugoslavia
(Vol. 43, no. 3, August 2005)
Find all sequences c = {ci }ni=1 , ci = ci (n) such that the inequality
|x∗ −
n
X
v
u n
n
X
uX
√
2
2
t
ci xi −
ci xi ,
ci xi | ≤ n − 1
i=1
i=1
i=1
holds for all sequences x = {xi }ni=1 of arbitrary real numbers and arbitrary x∗ ∈ x.
Solution by the proposer
We show that the conditions of the problem are satisfied if and only if ci = 1/n for
i = 1, . . . , n.
Putting xi = 1 P
for all i = 1,P. . . , n, we see that a necessary condition for the given
n
n
inequality to hold is ( i=1 ci )(1 − i=1 ci ) ≥ 0; i.e.,
0≤
n
X
ci ≤ 1.
(1)
i=1
Also, putting subsequently xi = 0, i 6= s, x∗s = xs = 1, for all s = 1, . . . , n, in the desired
inequality, we obtain
p
√
|1 − cs | ≤ n − 1 cs (1 − cs ), s = 1, . . . , n;
191
ADVANCED PROBLEMS AND SOLUTIONS
i.e, cs ≥ 1/n for all s = 1, . . . , n. But these last inequalities together with (1) imply that
cs = 1/n for all s = 1, . . . , n. This last condition is also sufficient since
n
n
Pn x2 Pn x 2 n − 1 X
X
2 2
i=1 i
i=1 i
(n − 1)
=
n
−
x
−
x
i
i
n
n
n2
i=1
i=1
n−1
=
n2
X
1≤i<j≤n
n
n
X
2
1
1 X
∗ 2
(xi − xj ) ≥ 2 (n − 1)
(xi − x ) ≥ 2
(xi − x∗ )
n
n i=1
i=1
2
=
Pn
i=1
xi
n
− x∗
2
,
which completes the proof.
Sums of three cubes
H-628 Proposed by Juan Pla, Paris, France
(Vol. 43, no. 3, August 2005)
Let us consider the set S of all the sequences {Un }n≥0 satisfying a second order linear
recurrence
Un+2 − aUn+1 + bUn = 0,
with both a and b rational integers, and having only integral values. Prove that for infinitely
many of these sequences their general term Un is a sum of three cubes of integers for any value
of the subscript n.
Solution by the proposer
The starting point is the following easy to prove identity
(x + y + z)3 − (x3 + y 3 + z 3 ) = 3(x + y)(y + z)(z + x).
(1)
Setting in (1): x = Un+2 , y = −aUn+1 , z = bUn , we obtain easily
3
Un+2
+ (−aUn+1 )3 + (bUn )3 = −3abUn+2 Un+1 Un .
(2)
But since we have the classical relation
2
Un+2 Un − Un+1
= bn (U2 U0 − U12 ),
after substitution of Un Un+2 in the right hand side of (2) and simplifications we obtain
3
3
Un+2
− (−3ab + a3 )Un+1
+ (bUn )3 = −3abn+1 (U2 U0 − U12 )Un+1 .
(3)
To have a sum of three cubes on the left hand side, we need only that −3ab+b3 be a cube, which
is the case if we set a to be an arbitrary integer and then look for b such that −3b + a2 = a2 c3 ,
or, equivalently, b = a2 (1 − c3 )/3, with an arbitrary integer c. In order for b to be an integer, it
suffices to impose that either a is a multiple of 3 or c ≡ 1 (mod 3). It is now easy to prove that
the sequence whose general term appears in the right hand side of relation (3) does belong to
the set S.
192